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# Water flow rate through an opening in a tube

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Please see attached file for the illustration.

Tube B (full of air) is sealed on the top and a 16" OD opening on the bottom. B is 104.6' long.
Container E (full of air) is inserted into section H and permanently fixed in place.
B is inserted fully into container C. Water rises 59.6' into B. H is 45' feet of air. See Illustration 1.

E is filled with water (see Illustration 2). E releases water through a 16" OD valve g while venting air through valve f.
E releases water at the rate of 100 gal per second for 8 seconds.

PROBLEM:
When E releases water at 100gal/s will water flow through h at the same rate?
Will the height of W change?

If E is supplied by a constant source of water , releases at a constant flow of 100 gal/s and container C remains at a constant depth will water flow through h at the same rate? Will the length of W remain the same?

https://brainmass.com/physics/gravity/water-flow-rate-through-opening-tube-335702

## SOLUTION This solution is FREE courtesy of BrainMass!

I however want to mention this: a couple of quantitative values were given in the question which kind of suggests quantitative answers too, but much more quantitative values are needed if the answers are to be purely quantitative; values like the cross-sectional area or diameter of tube B, the atmospheric pressure at the height were this experiment is performed, the value of acceleration due to gravity at that place, and the densities of water and air will be required to do necessary calculations, but the last statement in the question clearer suggests that a calculative answer is not what is needed.

So, I will help you address this question this way: answer the question in an essay form, giving you all the needed explanations, and presenting the physical & mathematical clues where necessary. (I will try to be as simple as possible). Feel free to neglect the mathematical aspects while making your own solution. I just included them for your savor in case you are interested. I have also marked them in bold so that you can easily see and skip them if you aren't interested.

SOLUTION:
First you will need to understand that since tube B has an open valve at h, the air pressure inside the tube just before it was inserted in the water of container C was equal to the value of the atmospheric pressure at h (that is, the value of atmospheric pressure at the surface of the water contained in container C).

Now, as tube B is inserted completely into container C, the air pressure inside tube B will increase by an amount that is equal to the value of pressure constituted by a column of water of height 104.6' (that is the length of tube B inserted in C). It is this additional pressure from the water that compresses the air in tube B into a 45' feet column and allowing water to rise 59.6' into B. This is where the value of pressure of air inside tube B becomes equal to the value at valve h. LET'S CALL THIS VALUE OF PRESSURE 'PNOW', it is the value of pressure at valve h so long as tube B remains in container C at this depth.

Note here that we could calculate the value of the additional pressure by using the formula:
Pressure=rho*g*h [where rho is the density of water, g is the value of acceleration due to gravity, and h is the height of the water column constituting the pressure (that is 104.6' in this case)].

Also, we could use Boyle's law to calculate the new air pressure inside B:
p1*v1=p2*v2 [p1 is the value of the air pressure in B before it was inserted in container C (that is the value of the atmospheric pressure at the surface of container C), p2 is the new value of air pressure in tube B after it has been completely inserted in water of container C (this is the value we wish to know), v1 is the volume of air in tube B before the insertion (this is the volume of B minus the volume of E), and v2 is the volume of air in tube B after the insertion(this is the value of v1 minus the volume of water that has risen 59.6' into B). We could deduce from the statement in the question "E releases water at the rate of 100 gal per second for 8 seconds" that the volume of E is 800 gal, this is also if we assume that all of the water in E was expelled in 8 seconds at the 100 gal per second rate]. BUT AS I MENTIONED EARLIER, I DO NOT THINK THE QUESTION REQUIRES US TO GO INTO ALL THESE CALCULATIONS.

Next as E releases water into B at a rate of 100 gal per second, empty spaces are automatically created in E at the same rate, and air from B vents into E through valve f at the same rate resulting in a further decrease in the length of air in tube B and an increase in the length of water in it.

Now to answer the first question "When E releases water at 100gal/s will water flow through h at the same rate?"
No, water won't flow through h at the same rate. There won't even be a net flow of water through h; if we neglect the speed with which water from E strikes the surface of tube B's water, then water won't flow through h at all, if however we don't neglect this speed, then the pressure generated by this strike causes some of the water in B to flow through h into container C at a rate determined by the speed with which E's water strikes the surface of B's water, but this outflow is quickly countered by an inflow of exactly the same amount of water from C through h into B (this happens to return the value of the pressure of air in B to 'PNOW', its original value before the drop), so there won't be a net flow of water through h either way. THIS IS BASICALLY BECAUSE THE FLOW OF WATER FROM E TO B DOESN'T CHANGE THE PRESSURE OF AIR IN B; THE AIR IS NOT COMPRESSED ANY FURTHER; AS WATER FROM E OCCUPIES SOME VOLUME OF B, AN EQUAL AMOUNT OF SPACE IS CREATED IN E FOR AIR IN B TO GO AND OCCUPY, SO THE AIR PRESSURE REMAINS UNCHANGED, AND THE AIR THEREFORE DOES NOT EXERT ANY ADDITIONAL PRESSURE THAT WILL CAUSE WATER TO FLOW OUT OF h.

Next the second question "Will the height of W change?" I guess you now know the answer to this question.
Yes, the height of W will change (actually increase) because water from E gets to settle in B, and not leaving through h. On the other hand, the height of H will decrease as the air in B enters E through f.

Finally, the last 2 questions "If E is supplied by a constant source of water , releases at a constant flow of 100 gal/s and container C remains at a constant depth will water flow through h at the same rate? Will the length of W remain the same?"

If E is supplied by a constant source of water, and keeps releasing at a constant flow of 100 gals/s, that means the water from E gets to occupy space in B and no space is created in E for B's air to vent into (similar to closing valve f), the result is that the pressure inside B tends to increase, but since container C remains at a constant depth (the pressure at valve h remains at 'PNOW', the original value), so the pressure that is trying to build inside B is directed to h with the net result being that water flows out of B through h at the same rate of 100 gal/s to keep the pressure inside B equal to 'PNOW'.

So, YES! Water will flow through h at the same rate.

Also, YES! The length of W will remain unchanged; this is because all of the water released into B from E gets to leave B through h at the same rate. So, the length of W remains the same.

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